Question

数学 求导

Answer 1

y = x^(2+cosπx)
lny = (2+cosπx)lnx
1/y * y′ = (2-πsinπx)lnx + (2+cosπx)/x
y ′ = x^(2+cosπx) * { (2-πsinπx)lnx + (2+cosπx)/x }

Answer 2

[x^(2+cosπx)]'
=x^(2+cosπx-1)(-sinπx)(π)
=-πsin(πx)x^(1+cosπx)